9/9/2004 sec 2_5 empty.doc 1/5 Jim Stiles The Univ. of Kansas Dept. of EECS 2-5 The Calculus of Scalar and Vector Fields (pp.33-55) Q: A: 1. 4. 2. 5. 3. 6. A. The Integration of Scalar and Vector Fields 1. The Line Integral9/9/2004 sec 2_5 empty.doc 2/5 Jim Stiles The Univ. of Kansas Dept. of EECS ()Ccrd⋅∫A A Q1: HO: Differential Displacement Vectors A1: HO: The Differential Displacement Vectors for Coordinate Systems Q2: A2: HO: The Line Ingtegral Q3: HO: The Contour C A3: HO: Line Integrals with Complex Contours Q4:9/9/2004 sec 2_5 empty.doc 3/5 Jim Stiles The Univ. of Kansas Dept. of EECS HO: Steps for Analyzing Line Integrals A4: Example: The Line Integral 2. The Surface Integral ()sSrds⋅∫∫A Q1: A1: HO: Differential Surface Vectors HO: The Differential Surface Vectors for Coordinate Systems Q2: A2: HO: The Surface Integral9/9/2004 sec 2_5 empty.doc 4/5 Jim Stiles The Univ. of Kansas Dept. of EECS Q3: HO: The Surface S A3: HO: Integrals with Complex Surfaces Q4: HO: Steps for Analyzing Surface Integrals A4: Example: The Surface Integral 3. The Volume Integral ()Vgrdv∫∫∫ Q1:9/9/2004 sec 2_5 empty.doc 5/5 Jim Stiles The Univ. of Kansas Dept. of EECS A1: HO: The Differential Volume Element HO: The Volume V Example: The Volume Integral9/7/2004 Differential Line Vector Elements.doc 1/4 Jim Stiles The Univ. of Kansas Dept. of EECS Differential Displacement Vectors The derivative of a position vector r, with respect to coordinate value A (where {},,,,,,xyz rρφθ∈A) is expressed as: ()()()()ˆˆˆˆˆˆˆˆˆxyzyxzxyzdr dxyzdddydx dzddddydx dzddd=++=++⎛⎞⎛⎞ ⎛⎞=++⎜⎟ ⎜⎟⎜⎟⎝⎠ ⎝⎠⎝⎠AAAAAAAAaaaaaaaaa A: The vector above describes the change in position vector r due to a change in coordinate variable A . This change in position vector is itself a vector, with both a magnitude and direction. Q: Immediately tell me what this incomprehensible result means or I shall be forced to pummel you !9/7/2004 Differential Line Vector Elements.doc 2/4 Jim Stiles The Univ. of Kansas Dept. of EECS For example, if a point moves such that its coordinate A changes from A to +∆AA, then the position vector that describes that point changes from r to + r∆A . In other words, this small vector ∆A is simply a directed distance between the point at coordinate A and its new location at coordinate +∆AA! This directed distance ∆A is related to the position vector derivative as: ˆˆˆ xyzdrddydx dzaaaddd∆=∆⎛⎞⎛⎞ ⎛⎞=∆ +∆ +∆⎜⎟ ⎜⎟⎜⎟⎝⎠ ⎝⎠⎝⎠AAAAAAAAA As an example, consider the case when ρ=A . Since cosxρφ= and sinyρφ= we find that: r + r∆A∆A9/7/2004 Differential Line Vector Elements.doc 3/4 Jim Stiles The Univ. of Kansas Dept. of EECS ()()ˆˆˆˆˆˆˆˆˆcos sincos sinxyzxyzxydydr dx dzaaadd d ddddzaaadddaaaρρρ ρ ρρφ ρφρρρφφ=++=++=+= A change in position from coordinates , , zρφ to , , zρρφ+∆ results in a change in the position vector from r to + r∆A . The vector ∆A is a directed distance extending from point ,,zρφ to point , , zρρφ+∆ , and is equal to: ˆˆˆrdcos sinxydaaaρρρρφ ρφρ∆=∆=∆ +∆=∆A If ∆A is really small (i.e., as it approaches zero) we can define something called a differential displacement vector dA : rˆpaρ∆=∆A y x9/7/2004 Differential Line Vector Elements.doc 4/4 Jim Stiles The Univ. of Kansas Dept. of EECS 00limlimddrddrdd∆→∆→∆⎛⎞=∆⎜⎟⎝⎠⎛⎞=⎜⎟⎝⎠AAA AAAAA For example: ˆrdddaddρρρρρ== Essentially, the differential line vector dA is the tiny directed distance formed when a point changes it location by some tiny amount, resulting in a change of one coordinate value A by an equally tiny (i.e., differential) amount dA. The directed distance between the original location (at coordinate value A) and its new location (at coordinate value d+AA) is the differential displacement vector dA. We will use the differential line vector when evaluating a line integral. r + rdA dA9/7/2004 The Differential Line Vector for Coordinate Systems.doc 1/3 Jim Stiles The Univ. of Kansas Dept. of EECS The Differential Displacement Vector for Coordinate Systems Let’s determine the differential displacement vectors for each coordinate of the Cartesian, cylindrical and spherical coordinate systems! Cartesian This is easy! ˆˆˆˆˆˆˆˆˆˆˆˆrrrxyzxxyzyxyzzdyddx dzdx dx dxdx dx dx dxdxdyddx dzdy dy dydy dy dy dydydyddx dzdz dz dzdz dz dz dzdz⎡⎤⎛⎞⎛⎞ ⎛⎞== + +⎢⎥⎜⎟ ⎜⎟⎜⎟⎝⎠ ⎝⎠⎝⎠⎣⎦=⎡⎤⎛⎞⎛⎞⎛⎞== + +⎢⎥⎜⎟⎜⎟⎜⎟⎢⎥⎝⎠⎝⎠⎝⎠⎣⎦=⎡⎤⎛⎞⎛⎞ ⎛⎞== + +⎢⎥⎜⎟ ⎜⎟⎜⎟⎝⎠ ⎝⎠⎝⎠⎣⎦=aaaaaaaaaaaa9/7/2004 The Differential Line Vector for Coordinate Systems.doc 2/3 Jim Stiles The Univ. of Kansas Dept. of EECS Cylindrical Likewise, recall from the last handout that: ˆdadρρρ= ˆˆˆˆrrrrxyzddx dx a dxdxddy dy a dydyddz dz a dzdzdddaddρρρρρ======== A: NO!! Do not make this mistake! For example, consider dφ: ()()ˆˆˆˆˆˆˆˆˆˆˆrcos sinsin ssin sxyzxyzxyxydddddydx dzaaadddddddzaaaddddacoada adco a dφφφφφφφφρφ ρφφφφφρφ ρ φφρφρφφ φ==++=++=− +=− + =⎛⎞⎜⎟⎝⎠⎛⎞⎜⎟⎝⎠Q: No!! ˆdadφφφρ=?!? How did the coordinate ρ get in there? Q: It seems very apparent that: ˆdad=AAA for all coordinates A; right ? Maria, look! I’m starting to see a trend!9/7/2004 The Differential Line Vector for Coordinate Systems.doc 3/3 Jim Stiles The Univ. of Kansas Dept. of EECS The scalar differential value dρφ makes sense! The differential displacement vector is a directed distance, thus the units of its magnitude must be distance (e.g., meters, feet). The differential value dφ has units of radians, but the differential value dρφ does have units of distance. The differential displacement vectors for the cylindrical coordinate system is therefore: ˆˆˆrrrpzdddadddddaddddz dz a dzdzφρρρρφφρφφ====== Likewise, for the spherical coordinate system, we find that: ˆˆˆrrrsinrddr dr a drdrdddardddddarddθφθθ θθφφθφφ======09/07/04 The Line Integral.doc 1/6 Jim Stiles The Univ. of Kansas Dept. of EECS The Line